Problem: $f\,^{\prime}(x)=3x^2-2x+7$ and $f(6)=200$. $f(1) = $
Solution: Finding $f(x)$ We have $f'(x)=3x^2-2x+7$ and we want to find $f(x)$ : $\begin{aligned}f(x) &= \int f'(x)\,dx \\\\ & = \int (3x^2-2x+7)\,dx \\\\ & = {x^3-x^2+7x} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(6)=200$. Here's what we get when we plug in $6$ : $\begin{aligned}f(6)&={(6)^3-(6)^2+7(6)} {+ C}\\\\ &={222} {+ C} \end{aligned}$ We are given that this must equal $200$ : $200 = {222} {+ C}$ Solving the equation gives us ${C=-22}$. Finding $f(1)$ Now, we have that $f(x)={x^3-x^2+7x} {-22}$. Let's find $f(1)$ by plugging in $1$ : $\begin{aligned}f(1)&=(1)^3-(1)^2+7(1) - 22\\\\ &=-15 \end{aligned}$ The answer $f(1) = -15$